| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Topic | Inequalities |
| Type | Graph feasible region from inequalities |
| Difficulty | Standard +0.8 This question requires algebraic manipulation to derive a quadratic in x, careful analysis of the discriminant to establish the constraint y ≤ 1, then using the explicit formula to prove an inequality xy < 2. The multi-step reasoning connecting algebra to inequalities and geometric interpretation is more sophisticated than standard A-level fare, though each individual technique is accessible. The constraint analysis and inequality proof require insight beyond routine manipulation. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02g Inequalities: linear and quadratic in single variable1.02i Represent inequalities: graphically on coordinate plane |
| Answer | Marks | Guidance |
|---|---|---|
| Combine and simplify the two fractions: \(y = \frac{x - i + x + i}{x^2 - i^2} - \frac{2x}{x^2 + 1}\) | B1 | |
| Complete attempt to solve quadratic \(yx^2 - 2x + y = 0\) for \(x\) | M1 | |
| Obtain \(x = \frac{2 \pm \sqrt{4 - 4y^2}}{2y} = \frac{1 \pm \sqrt{1 - y^2}}{y}\) | A1 AG | |
| State that the reality of \(x \Rightarrow\) positivity of the discriminant and that the positivity of \(y \Rightarrow y \leq 1\) | A1 AG | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| The larger solution for \(x\) satisfies \(xy = 1 + \sqrt{(1-y^2)}\) | M1 | |
| \(\leq 2\) because \(0 < y \leq 1\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Sketch of region in the first quadrant, bounded by the horizontal line \(y = 1\) (region below) and the rectangular hyperbola \(xy = 2\) (curve acceptable) | B1 B1 B1 | [3] |
### (i)
Combine and simplify the two fractions: $y = \frac{x - i + x + i}{x^2 - i^2} - \frac{2x}{x^2 + 1}$ | B1 |
Complete attempt to solve quadratic $yx^2 - 2x + y = 0$ for $x$ | M1 |
Obtain $x = \frac{2 \pm \sqrt{4 - 4y^2}}{2y} = \frac{1 \pm \sqrt{1 - y^2}}{y}$ | A1 AG |
State that the reality of $x \Rightarrow$ positivity of the discriminant and that the positivity of $y \Rightarrow y \leq 1$ | A1 AG | [4]
### (ii)
The larger solution for $x$ satisfies $xy = 1 + \sqrt{(1-y^2)}$ | M1 |
$\leq 2$ because $0 < y \leq 1$ | A1 | [2]
*Note: There are acceptable alternative methods.*
### (iii)
Sketch of region in the first quadrant, bounded by the horizontal line $y = 1$ (region below) and the rectangular hyperbola $xy = 2$ (curve acceptable) | B1 B1 B1 | [3]It is given that
$$y = \frac{1}{x+1} + \frac{1}{x-1},$$
where $x$ and $y$ are real and positive, and $i^2 = -1$.
\begin{enumerate}[label=(\roman*)]
\item Show that
$$x = \frac{1 \pm \sqrt{1-y^2}}{y} \quad \text{and} \quad y \leqslant 1.$$ [4]
\item Deduce that
$$xy < 2.$$ [2]
\item Indicate the region in the $x$-$y$ plane defined by
$$y \leqslant 1 \quad \text{and} \quad xy < 2.$$ [3]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2010 Q5 [9]}}